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Finite Element Truss Problem

6.0 Trusses Using FEAWe started this series of lectures looking at truss problems. We limited the

discussion to statically determinate structures and solved for the forces in elements andreactions at supports using basic concepts from statics.

Next, we developed some basic one dimensional finite elements concepts bylooking at springs. We developed a basic system of node numbering that allows us to

solve problems involving several springs by simply adding the stiffness matrices for eachspring.

In the next two lectures, we will extend the basic one dimensional finite elementdevelopment to allow us to solve generalized truss problems. The technique we willdevelop is a little more complex than that originally used to solve truss problems, but it

allows us to solve problems involving statically indeterminate structures.

6.1 Local and Global CoordinatesWe can extend the one dimensional finite element analysis by looking at a one

dimensional problem in a two dimensional space. Below we have a finite element (couldbe a spring) that attaches to nodes 1 and 2.

The y, coordinates are the local coordinates for the element and y, are the globalcoordinates. The local coordinate system looks much like the one dimensional

coordinate system we developed in the last lecture.We can convert the displacements shown in the local coordinate system by

looking at the following diagram. We will let 1q and 2q represent displacements in thelocal coordinate system and q1, q2, q3, and q4 represent displacements in the x-y (global)

1

2

x

yLocal coordinate

system

x

y

Global coordinate

System

Figure 1 - Local and global coordinate systems

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coordinate system. Note that the odd subscripted displacements are in the x direction andthe even ones are in the y direction as shown in the following diagram.

In a previous lecture we looked at the deformation of springs by looking at thedisplacements at the ends of the springs. Here we are going a step farther into finiteelement development by looking at the strain energy of the element. The element could

be a spring but in this case we will generalize and look at it as any solid material element.

The only restriction we will place on the element is that the deformation is smallcompared to its total length.

We know from Hooks law that the force is directly proportional to thedeformation.

xkF = (6.1)We can compute the energy by integrating over the deformation

2

02

1kQxdxku

Q

== (6.2)

whereL

AEk = the element stiffness, A = the cross sectional area of the element,

E = Youngs modulus for the material, and L = the length of the element. Q is the totalchange in length of the element. Note that we are assuming the deformation is linear

over the element. All equal length segments of the element will deform the sameamount. We call this a constant strain deformation of the element.

We can rewrite this change in length as

q1

q2

?

Deformed element

sin2q

q1

q2

q4

q3

cos1q

Un-deformed element

Figure 2 - The deformation of an element in both local and global coordinate systems.

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)( '1'

2 qqQ = (6.3)

Substituting this into equation (6.2) gives us

212 )(

21 qqku = (6.4)

or

)2(2

1 2112

2

2 qqqqku += (6.5)

Rewriting this in vector form we let

=

2

1

q

qq (6.6)

and

=

11

11

L

AEk (6.7)

With this we can rewrite equation (6.5) as:

qkqu T =2

1(6.8)

We can do the indicated operations in (6.8) to see how the vector notation works.We do this by first expanding the terms then doing the multiplication.

{ }

=

2

1

2111

11

2 q

qqq

L

AEu (6.9)

{ }

+=

2

1

21212 q

qqqqq

L

AEu (6.10)

( ))()(2

122211 qqqqqqL

AEu += (6.11)

( )212

221

2

12

qqqqqqL

AEu += (6.12)

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)2(2

2

221

2

1 qqqqL

AEu += (6.13)

Which is the same as equation (6.5).

Equation (6.7) is the stiffness matrix for a one dimensional problem. It bears veryclose resemblance to equation (5.7) used in our one dimensional spring development.

[ ]

=

kk

kkK (5.7)

=

11

11

L

AEk (6.7)

6.2 Two Dimensional Stiffness MatrixWe know for local coordinates that

=

2

1

q

qq (6.6)

and for global coordinates (See Figure 2)

=

4

3

2

1

q

q

q

q

q (6.14)

We can transform the global coordinates to local coordinates with the equations

sincos 211 qqq += (6.15)and

sincos432qqq += (6.16)

This can be rewritten in vector notation as:

qq = (6.17)

where

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=

sc

scM

00

00, (6.18)

cos=c , and sin=s .

Using

qkqu T =2

1(6.8)

we can substitute in equation (6.17)

[ ]qMkMqu TT =2

1(6.19)

Now we will let

MkMk T = (6.20)

and doing the multiplication, k our stiffness matrix for global two dimensionalcoordinates becomes

=

22

22

22

22

scsscs

csccsc

scsscs

csccsc

L

AEk (6.21)

where:

E = Youngs modulus for the element materialA = the cross sectional area of the element

L = the length of the element

cos=c sin=s

6.3 Stress ComputationsThe stress can be written as

E= (6.22)

where is the strain, the change in length per unit of length. We can rewrite this as:

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L

qqE 12

= (6.23)

In vector form we can write the equation as

{ }

=

2

111

q

q

L

E (6.24)

From our previous discussion, we know that in local coordinates

=

2

1

q

qq (6.25)

and in global coordinates

=

4

3

2

1

q

q

q

q

q (6.26)

From equation (6.17) we know that

qq = (6.17)

where

=

sc

scM

00

00(6.18)

Substituting this in to the equation (6.24) yields

{ } qML

E11= (6.27)

Now we multiply M by the vector

{ }qscscL

E= (6.28)

total deformation

length of element